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Math Help - Determining straight line characteristics for 1st order PDE's

  1. #1
    Senior Member bugatti79's Avatar
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    Determining straight line characteristics for 1st order PDE's

    Dear Folks,

    Calculate the straight line characteristics for the following linear 1st order PDE's

    1)  3u_x +5u_y -xyu=0

    I calculate k=y-\frac{5}{3}x and when transformed becomes \eta=y-\frac{5}{3}\xi

    but the book its given as 5x-3y=k and when transformed becomes 5\xi-3y=\eta

    2) yu_x+x^2uy=xy

    I calculate k=-\frac{x^3}{3}+\frac{y^2}{2} when transformed becomes \eta=\frac{-\xi^3}{3}+\frac{y^2}2}

    but the book is k=2x^3-3y^2 when transformed becomes \eta=2\xi^3-3y^2
    The only way I can see them arriving at this is k'=\frac{-x^3}{3}+\frac{y^2}{2} therefore
    -6k'=2x^3-3y^2=k

    The pattern I see in these kind of questions is that when k is written in terms of x and y that the 'x' terms are always positve before transforming. Why is this?

    Thanks
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  2. #2
    MHF Contributor
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    It all comes down to a matter of prefrence. The writter of the solutions (I addition to myself) doesn't like fractions and prefers x before y.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    It all comes down to a matter of prefrence. The writter of the solutions (I addition to myself) doesn't like fractions and prefers x before y.
    so its ok to change the sign so long as its accounted for in the constant -6k'=k etc as above? Ie, k=-\frac{x^3}{3}+\frac{y^2}{2} becomes k=2x^3-3y^2

    Thanks
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  4. #4
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    Let me show you by example.

    Consider your example

    y u_x + x^2 u_y = xy

    Let me choose

    (1) r = - \dfrac{x^3}{3} + \dfrac{y^2}{2},\;\;\; s = x

    (2) r =  \dfrac{x^3}{3} - \dfrac{y^2}{2},\;\;\; s = x

    (3) r = 2x^3 - 3y^2,\;\;\; s = x.

    All three transform your PDE to

    u_s = s\;\;\; \text{so}\;\;\; u = \dfrac{s^2}{2} + F(r).

    Thus, your solution is

    (1) u = \dfrac{x^2}{2} + F\left( - \dfrac{x^3}{3} + \dfrac{y^2}{2}\right)

    (2) u = \dfrac{x^2}{2} + F\left( \dfrac{x^3}{3} - \dfrac{y^2}{2}\right)

    (3) u = \dfrac{x^2}{2} + F\left(2x^3 - 3y^2 \right).

    As F is arbitrary you can always choose F(\lambda) = F(k \cdot \lambda), thus making all three equivalent.
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  5. #5
    Senior Member bugatti79's Avatar
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    As F is arbitrary you can always choose F(\lambda) = F(k \cdot \lambda), thus making all three equivalent.
    so F(-6(\frac{-x^3}{3}+\frac{y^2}{2}))=F(2x^3-3y^2) where \lambda=-6

    Im happy with that. Thanks Danny
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  6. #6
    Senior Member bugatti79's Avatar
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    sorry, that should be k=-6
    cheers
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  7. #7
    MHF Contributor
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    What I really should have said to be consistent with your notation is

    F(k) = F(\lambda k).
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